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katrin [286]
3 years ago
7

Supercharging is the process of (a) Supplying the intake of an engine with air at a density greater than the density of the surr

ounding atmosphere (b) providing forced cooling air (e) injecting excess fuel for raising more load (d) supplying compressed air to remove combustion products fully. (e) raising exhaust pressure.
Engineering
1 answer:
iVinArrow [24]3 years ago
4 0

Answer:

a)supplying the  intake of an engine  with air at a  density greater  than the density  of the surrounding  atmosphere

Explanation:

Supercharging  is the process of  supplying the  intake of an engine  with air at a  density greater  than the density  of the surrounding  atmosphere.

By doing this , it increases  the power out put  and increases the  brake thermal  efficiency of the  engine.It also  increases the  volumetric efficiency of the  engine.

So the our  option a is  right.

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Select three types of lines that engineers use to help represent the shape of a design in a sketch.
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Hidden lines

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Extension lines:-

  • Used to explain the expansion of structures like building

Object lines

  • Used to describe the structure of objects and the lining to show borders
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2 years ago
25 points and brainliest is it A, B, C, D
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2 years ago
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A compound machine contains three simple machines with IMAs of 2, 4 and 5, respectively. What is the overall ideal mechanical ad
anygoal [31]

Answer:

Overall ideal mechanical advantage of the machine = 40

Explanation:

Given:

Ideal mechanical advantage of three machine = 2, 4, 5

Find:

Overall ideal mechanical advantage of the machine

Computation:

Overall ideal mechanical advantage of the machine = 2 × 4× 5

Overall ideal mechanical advantage of the machine = 40

3 0
2 years ago
Additional scals apply to the
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2 years ago
The production process of rods from machine "A" yields specimen with the following specs. Mean: µ(LA)=20.00mm, STD: s(LA)=0.50mm
Oxana [17]

Answer: the standard deviation STD of machine B is s (Lb) = 0.4557

Explanation:

from the given data, machine A and machine B produce half of the rods

Lt = 0.5La + 0.5Lb

so

s² (Lt) = 0.5²s²(La) + 0.5²s²(Lb) + 0.5²(2)Cov (La, Lb)

but Cov (La, Lb) = Corr(La, Lb) s(La) s(Lb) = 0.4s (La) s(Lb)

so we substitute

s²(Lt) = 0.25s² (La) + 0.25s² (Lb) + 0.4s (La) s(Lb)

0.4² = 0.25 (0.5²) + 0.25s² (Lb) + (0.5)0.4(0.5) s(Lb)

0.64 = 0.25 + s²(Lb) + 0.4s(Lb)

s²(Lb) + 0.4s(Lb) - 0.39 = 0

s(Lb) = { -0.4 ± √(0.16 + (4*0.39)) } / 2

s (Lb) = 0.4557

therefore the standard deviation STD of machine B is s (Lb) = 0.4557

8 0
2 years ago
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